concentration of solutions (molality, molarity)

8/11/2019 Concentration of Solutions (Molality, Molarity)

1/5

Chapter 10: Solutions

10. Refer to Section 10.1 and Examples 10.4 and 10.5.

Molality

Mass Percent

of Solvent Ppm Solute

Mole Fraction

of Solvent

a. 2.577 86.58% 1.340 x105 0.9556

b. 20.4 45.0 5.50 x105 0.731

c. 0.07977 99.5232% 4768 0.9986

d. 12.6 57.0% 4.30 x105 0.815

Since these are aqueous solutions, the solvent is water. The solute is urea, CO(NH2)2.

Molar mass of urea: (12.01) + 16.00 + 2(14.01 + 2(1.008)) = 60.06 g/mol.

a. 2.577 mindicates that there are 2.577 mol. of solute per 1000 g of solvent and the

calculations are based on that ratio.

ureag154.8=ureamol.1

ureag60.06mol.2.577 x

OHmol.55.49=OHg18.02

OHmol.1

kg1

g1000kg1 2

2

2xx

86.58%=g1155

g1000=100%g)154.8+g(1000

g1000=%mass x

5101.340 x=10g)154.8+g(1000

g154.8=soluteppm 6x

0.9556=mol.)2.577+mol.5.495(

mol.5.495=OH2X


2/5

b. 45.0 mass % of solvent means that 45.0 g of solvent are present for each 100.0 g of

solution. Calculations are then based on this ratio.

OHmol.2.50=g18.02

OHmol.1g45.0 2

2x

mass of solute = 100.0 g (total) 45.0 g (water) = 55.0 g CO(NH2)2

ureamol.0.916=ureag60.06

ureamol.1ureag55.0 x

m20.4=OHkg0.0450

ureamol.0.916=molality

2

(remember to convert mass of solvent to kg)

5105.50 x=10

g100

ureag55.0=soluteppm 6x

0.731=mol.)0.916+mol..502(

mol..502=X OH2

c. g7684=10solutionofmass

soluteofmass=ppm 6x

If we assume a total mass of 106g, then the mass of solute = 4768 g by definition. Any

assumption for mass is valid here, this one was chosen for simplicity.

ureamol.9.397=ureag60.06

ureamol.1ureag4768 x

OHmol.5244.05=g18.0152

mol.1OHg952329g;952329=g4768-g10 22

6x

0.9986=mol.)79.39+mol.5244.05(

mol.5244.05=X OH2

99.5232%=100%g10

g109.95232=%mass

6

5

xx

m=kg995.232

mol.79.39=molality 0.07977


3/5

d. Xsolvent= 0.815 indicates that there are 0.815 moles H2O per 1 mole of solvent and solute

combined. Consequently, there must be 0.185 mol. (1-0.815) urea.

ureag11.1=ureamol.1

ureag60.06ureamol.0.185 x

OHg14.7=mol.1

g18.02OHmol.0.815 22 x

m12.6=OHkg0.0147

ureamol.0.185

2

(remember to convert mass of solvent to kg)

57.0%=100%g)14.7+g(11.1

g14.7=100%

masstotal

solventofmass=%Mass xx

5104.30 x=10

g)14.7+g(11.1

ureag11.1=10

solutionofmass

soluteofmass=ppm 66 xx

20. Refer to Sections 10.2 and 9.3.

The compound which exhibits intermolecular forces most similar to water will be the more

soluble in water (like dissolves like). Recall that water has dispersion, dipole, and H-bonding

forces.

a. CH3Cl: dispersion forces

CH3OH: dispersion, dipole, and H-bonding forces

CH3OH would be more soluble since it shares H-bonding with water.

b. NI3: dispersion and dipole forces

KI: ionic

KI would be more soluble because it is ionic and ionic compounds generally exhibit high

solubility in water.

c. LiCl: ionic

C2H5Cl: dispersion and dipole forces

LiCl would be more soluble because it is ionic and ionic compounds generally exhibit

high solubility in water.

d. NH3: dispersion, dipole, and H-bonding forces

CH4: dispersion and dipole forcesNH3would be more soluble because of the H-bonding.


4/5

30. Refer to Section 10.3.

The first step is to calculate the mole fraction of oxalic acid.

Hgmm0.41Hgmm21.97-Hgmm22.38- solutionOH 2 === PPP o

Hg)mm22.38)(Hgmm0.41))((4222422 OCHOHOCH

(XPXP == o

0.018422 OCH =X

Assuming 1 mole total, this means we have 0.018 mol. H2C2O4and 0.982 mol. water

(1.00 - 0.018 = 0.982). The next step is to calculate the masses associated with these

quantities, and from that the mass of solution and volume of solution.

OHg17.7=OHmol.1

OHg18.02OHmol.0.982

22

2

2

x

422

422

422422 OCHg1.6=

OCHmol.1

OCHg90.04OCHmol.0.018 x

L0.0184=mL1000

L1

g1.05

mL1solutiong1.6)+(17.7 xx

Now one can either:

1. Calculate molarity (mol./L) and convert mol./L to grams/L (using molecular mass) or

2. Directly calculate grams of H2C2O4in one liter (as shown below)

422422 OCHg87=

solutionL0.0184

OCHg1.6solutionL1.00 x

Thus, to prepare the prescribed solution, one must dissolve 87 g H2C2O4in enough water to

make 1.00 L of solution.


5/5

40. Refer to Sections 10.3 and 3.3, and Example 10.7.

Use the freezing point depression to calculate molality, and from that, moles of the compound

and the compound's molecular mass.

Tf= Tf -Tf= 178.40C 173.44C = 4.96C

Tf= kfm

4.96C = (40.0C/m)(m)

m= 0.124 m

camphorkg.05000=g1000

kg1camphorg50.00 x

camphorkg05000.0

mol.0.124

solventkg

solutemol.

xmm ==

x= 0.00620 mol.

g/mol.403mol.0.00620

g2.50=

concentration of solutions (molality, molarity)

Documents